Since ordering within (a,b) and (c,d), respectively, doesn't matter, let's let a <= b and c <= d. Assume that a != c. Then we'd need distinct pairs of integers in the range [2,9] with the same product. We can see by associating prime factors that the only possibilities are:

4 * 3 = 6 * 24 * 4 = 8 * 26 * 3 = 9 * 26 * 4 = 8 * 36 * 6 = 9 * 4

In none of those cases do the four digits sum to their product. Thus a = c and b = d.

So now we have 2*a + 2*b = a*b.

If a = b, then we have 4*a = a*a, so a = 4 and the solution is 4444. 4 is an upper bound on a since we assumed a < b and if they are both > 4 then the product will exceed twice their sum.

If a = 3, then we have 6 + 2*b = 3*b so b = 6, and solutions are 3636, 3663, 6336, and 6363.

If a = 2 then we have 4 + 2*b = 2*b so there are no additional valid solutions.